Proof of floor function is monotone

Question

Can anyone give a hint to approach this question? I have no idea

so far,

$f(x_1) = \lfloor x_1 \rfloor = n_1$ where ; $n_1 \leq x_1 < n_1 + 1$

$f(x_2) = \lfloor x_2 \rfloor = n_2$ where ; $n_2 \leq x_2 < n_2 + 1$

$x_1 < x_2 $ should imply $f(x_1)<f(x_2)$ (for increasing)

Answer 1

With what you have so far, I would try a direct proof.

Note: The definition of monotonic function is: If for all $x$, $y$ such that $x \geq y$, then $f(x) \geq f(y)$.

The definition you stated is for strictly increasing.

Let $x_1, x_2 \in \mathbb{R}$. Suppose $x_1 \leq x_2$, then $f(x_1) = \lfloor x_1 \rfloor = n_1 \leq x_1 \leq x_2 < n_2+1$. This implies $x_1 <n_2 +1 = \lfloor x_2 \rfloor + 1 = f(x_2) +1$

so we have $f(x_1) \leq x_1 \leq n_2 +1 <f(x_2) + 1$.

In conclusion, we now have $f(x_1)<f(x_2)+1$. We know that $f(x_1), f(x_2)$ are both integers, so what conclusion can we make about $f(x_1)$ and $f(x_2)$ ?